CODEWITHSUNDEEP
pythonstringintextract
Kamikaze_goldfish
Kamikaze_goldfish

Reputation: 861

Extract number from string in python without re

How can I extract a number from a string in python without having to use regex? I have seen isinstance but the number could change to almost anything. Any ideas?

https://www.investopedia.com/articles/retirement/?page=6

Upvotes: 1

Views: 2632

Answers (5)

yoonghm
yoonghm

Reputation: 4625

I know you do not need re, but it is actually very powerful. Under the hood, most libraries make use of re. Here is my solution to handle this situation:

import re

url = "www.fake888.com/article/?article=123&page=9&group=8"

numbers = re.findall(r'(?<==)(\d+)', url)
print(f'Found: {" ".join(numbers)}')

varval = re.findall(r'(\w+)=(\d+)', url)
urldict = {}
for var in varval:
  urldict[var[0]] = var[1]

print(urldict)

The output is

Found: 123 9 8
{'article': '123', 'page': '9', 'group': '8'}

Upvotes: 0

yvesva
yvesva

Reputation: 760

This assumes that there isn't multiple blocks of integers (e.g. www.something212.com/page=?13)

You could try using list comprehensions and str.isdigit()

url = 'https://www.investopedia.com/articles/retirement/?page=6'

digits = [d for d in url if d.isdigit()]

digit = ''.join(digits)

digit
>>> 6

Edited: now works with digits above 9

Upvotes: 1

Dani Mesejo
Dani Mesejo

Reputation: 61910

You can extract continuous groups of digits, anywhere on the string, using the following:

from itertools import groupby

url = 'https://www.investopedia.com/articles/retirement/?page=6&limit=10&offset=15'
print([int(''.join(group)) for key, group in groupby(iterable=url, key=lambda e: e.isdigit()) if key])

Output

[6, 10, 15]

Upvotes: 2

vash_the_stampede
vash_the_stampede

Reputation: 4606

If the url always has that format with only digits at the end you could do this: 

s = 'https://www.investopedia.com/articles/retirement/?page=25'
new = []
k = list(s)
[new.append(i) for i in k if i.isdigit()]
print(''.join(new))

(xenial)vash@localhost:~/python/stack_overflow$ python3.7 isdigit.py
25

Upvotes: 1

wim
wim

Reputation: 362726

It's a bit verbose, but I would use url parsing for this. The advantage overy regex is that you would get some input validation for free, and more readable code.

>>> from urllib.parse import urlparse, parse_qs
>>> url = 'https://www.investopedia.com/articles/retirement/?page=6'
>>> parsed = urlparse(url)
>>> query = parse_qs(parsed.query)
>>> [page] = query['page']
>>> int(page)
6

Upvotes: 2

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